Mathematical Methods for Physics and Engineering - Matematica.NET

15.4 EXERCISES15.3.6 Equations having y = Ae x as a solutionFinally, we note that if any general (linear or non-linear) nth-order ODE issatisfied identically by assuming thaty = dydx = ···= dn ydx n (15.88)then y = Ae x is a solution of that equation. This must be so because y = Ae x isa non-zero function that satisfies (15.88).◮Find a solution of(x 2 + x) dy d 2 ydx dx − 2 x2 y dy ( ) 2 dydx − x =0. (15.89)dxSetting y = dy/dx = d 2 y/dx 2 in (15.89), we obtain(x 2 + x)y 2 − x 2 y 2 − xy 2 =0,which is satisfied identically. Therefore y = Ae x is a solution of (15.89); this is easilyverified by directly substituting y = Ae x into (15.89). ◭Solution method. If the equation is satisfied identically by making the substitutionsy = dy/dx = ···= d n y/dx n then y = Ae x is a solution.15.4 Exercises15.1 A simple harmonic oscillator, of mass m and natural frequency ω 0 , experiencesan oscillating driving force f(t) =ma cos ωt. Therefore, its equation of motion isd 2 xdt + 2 ω2 0x = a cos ωt,where x is its position. Given that at t = 0 we have x = dx/dt = 0, find thefunction x(t). Describe the solution if ω is approximately, but not exactly, equalto ω 0 .15.2 Find the roots of the auxiliary equation for the following. Hence solve them forthe boundary conditions stated.(a) d2 fdt +2df2 dt +5f =0, with f(0) = 1,f′ (0) = 0.(b) d2 fdt +2df2 dt +5f = e−t cos 3t, with f(0) = 0,f ′ (0) = 0.15.3 The theory of bent beams shows that at any point in the beam the ‘bendingmoment’ is given by K/ρ,whereK is a constant (that depends upon the beammaterial and cross-sectional shape) and ρ is the radius of curvature at that point.Consider a light beam of length L whose ends, x =0andx = L, are supportedat the same vertical height and which has a weight W suspended from its centre.Verify that at any point x (0 ≤ x ≤ L/2 for definiteness) the net magnitude ofthe bending moment (bending moment = force × perpendicular distance) due tothe weight and support reactions, evaluated on either side of x, isWx/2.523